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The classical Abel's theorem for curves states that the fiber of Abel-Jacobi map $$Sym^kC\to J(C),\ (x_1,\ldots,x_k)\mapsto \sum_{i=1}^k\int_{x_0}^{x_i}$$ as a linear functional on $H^0(C,\Omega_C)$ modulo period, is the rational equivalent classes of the divisor $x_1+\cdots+x_k$. In particular the fiber is a projective space, and when $k$ is large, it is a projective bundle.

I'm wondering if it is known for cubic threefold, the meaning and geometric model of fiber of albanese map?

Let $X$ be a smooth cubic threefold, $F$ the Fano surface of lines. By Clemens and Griffiths, the Albanese variety of $X$ is isomorphic to the intermediate Jacobian. So we can consider the same "summation Albanese map"

$$\phi_k:Sym^kF\to Alb(X)\cong J(X), \ (x_1,\ldots,x_k)\mapsto \sum_{i=1}^k\int_{x_0}^{x_i}$$

which is again integrating holomorphic 1-forms. (Note the usual Abel-Jacobi map is given by "subtraction", which is different from here.)

Question: What is the fiber of $\phi_k$?

Note when $k=1$, $\phi_1$ is an embedding [Beauville 1982]. When $k=2$, $\phi_2$ is generically one-to-one onto its image according to [Beauville 2000, Remark 6.5].

Also, symmetric product of $F$ is singular, one may replace it by Hilbert scheme, but I just want to what is a general fiber, say birationally.

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    $\begingroup$ Please confer Arapura-Archava. $\endgroup$ Commented Jun 23 at 13:10
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    $\begingroup$ Thanks for pointing it out! Arapura-Archava implies the symmetric product $Sym^kF$ is not uniruled. In particular, the general fiber cannot be a projective space (at least) when $\phi_k$ is dominant. $\endgroup$
    – AG learner
    Commented Jun 23 at 18:21

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